IA-automorphism group of nilpotent group equals stability group of lower central series

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Statement

For an individual automorphism

Suppose G is a nilpotent group. Then, the following are equivalent for an automorphism \sigma of G:

  1. \sigma is an IA-automorphism of G, i.e., it induces the identity map on the abelianization of G.
  2. \sigma is a stability automorphism for the lower central series of G.

Note that G being nilpotent is important only in so far as it guarantees that the lower central series reaches the trivial subgroup. A slight variant of the statement would be true for non-nilpotent groups, but we wouldn't use the jargon of stability automorphism.

For subgroups of the automorphism group

Suppose G is a nilpotent group. The following subgroups of the automorphism group of G are equal:

  1. The subgroup of IA-automorphisms of G, i.e., automorphisms that induce the identity map on the abelianization of G.
  2. The stability group (i.e., the group of stability automorphisms) for the lower central series of G.

Proof

Background fact on iterated commutator mapping

The proof basically follows from the fact that for any positive integer n, the left-normed iterated commutator gives a surjective n-linear map from the abelianization of G to the quotient group \gamma_n(G)/\gamma_{n+1}(G) between successive members of the lower central series. Here, \gamma_n(G) is the n^{th} member of the lower central series of G. Explicitly, it is a n-linear map of abelian groups:

G/G' \times G/G' \times \dots G/G' \to \gamma_n(G)/\gamma_{n+1}(G)

This mapping is canonical, hence covariant with automorphisms.

(1) implies (2)

Given: An IA-automorphism \sigma of a nilpotent group G.

To prove: \sigma induces the identity map on each of the successive quotients of the lower central series, i.e., for every n, \sigma induces the identity map on each of the groups of the form \gamma_n(G)/\gamma_{n+1}(G) where \gamma_n(G) is the n^{th} member of the lower central series of G.

Proof: The covariance of the iterated commutator mapping with respect to \sigma, along with its surjectivity to \gamma_n(G)/\gamma_{n+1}(G), guarantees that since \sigma fixes G/G' pointwise, it also fixes the image \gamma_n(G)/\gamma_{n+1}(G) pointwise.

(2) implies (1)

Given: A nilpotent group G. An automorphism \sigma induces the identity map on each of the successive quotients of the lower central series, i.e., for every n, \sigma induces the identity map on each of the groups of the form \gamma_n(G)/\gamma_{n+1}(G) where \gamma_n(G) is the n^{th} member of the lower central series of G.

To prove: \sigma is an IA-automorphism of G.

Proof: This follows directly from setting n = 1.